Problem 10711

(American Mathematical Monthly, Vol.106, February 1999) Proposed by F. Luca (Germany).

A natural number isperfect if it is the sum of its proper divisors. Prove that two consecutive num- bers cannot both be perfect.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Firenze, viale Morgagni 67/A, 50134 Firenze, Italy.

We denote by σ(n) the sum of the divisors of the natural number n. It is easy to see that the arithmetic function σ is multiplicative that is

σ(a · b) = σ(a) · σ(b) iff (a, b) = 1.

Then n is a perfect number if and only if σ(n) = 2n. Moreover the following two elementary properties of perfect numbers hold:

(i) if n is even then n is perfect if and only if

n= 2^{q−1}(2^{q}−1)
where q and 2^{q}−1 are primes.

(ii) if n is an odd perfect number then it must have the form
n= p^{4m+1}s^{2}
where p is a prime such that [p]4= 1 and (p, s) = 1.

For a proof of the above results see for example L. Dickson, History of the theory of numbers, vol.

1, 1919, pp. 1-33. Note that the question as to whether there exist odd perfect numbers is as yet unsolved.

Therefore, by (i), two consecutive numbers are both perfect if

n_{−}= 2^{q−1}(2^{q}−1) − 1 or n+= 2^{q−1}(2^{q}−1) + 1

is perfect. The prime q is not 2 because neither n_{−}= 5 nor n+= 7 is perfect. Hence q = 2a + 1 ≥ 3.

1) Assume that n_{−} is perfect. Then

[n_{−}]4= [4^{a}]4·[2 · 4^{a}−1]4+ [−1]4= [3]4.

By (ii), n_{−} = p^{4m+1}s^{2} with [p]4 = 1. Since s is odd then either [s]4 = 1 or [s]4 = 3 and in
both cases [s^{2}]4= 1. So we have the contradiction

[3]4= [n_{−}]4= [p]4^{4m+1}·[s^{2}]4= [1]4.
2) Assume that n+ is perfect. Then

[n+]3= [4^{a}]3·[2 · 4^{a}−1]3+ [1]3= [1]3+ [1]3= [2]3.

By (ii), n+= p^{4m+1}s^{2} where p is a prime such that (p, s) = 1. Now, it easy to verify that
[p^{4m+1}]3·[s^{2}]3= [2]3 iff [p]3= [2]3 and [s^{2}]3= 1.

Moreover, since p is prime,

[σ(p^{4m+1})]3= [1 + p + . . . + p^{4m}+ p^{4m+1}]3= [1 + 2 + . . . + 1 + 2]3= [0]3.
By the multiplicative property of σ, we have the contradiction

[1]3= [2n+]3= [σ(n+)]3= [σ(p^{4m+1})]3·[σ(s^{2})]3= [0]3.